The eigen vector $(s)$ of the matrix $$\begin{bmatrix} 0 &0 &\alpha\\ 0 &0 &0\\ 0 &0 &0 \end{bmatrix},\alpha \neq 0$$ is (are)

- $(0,0,\alpha)$
- $(\alpha,0,0)$
- $(0,0,1)$
- $(0,\alpha,0)$

### 7 Comments

Here we got eign values as 0,0,0.

Now to get eign vector take λ=0 in

( A-λI) X =0 or AX=0 -----(i)

here X is eign vector in form of column matrix [X1,X2,X3].

If we multiply as per (i) we get equation as

X1*0+X2*0+ aX3=0

Now for LHS to be 0 , X3 must be 0 . Here X1 and X2 can be assumed any value . As per the given option a and c are wrong.hence b and d satisfy the case

## 6 Answers

So the question asks

$\begin{bmatrix} 0 &0 &\alpha \\ 0&0 &0 \\ 0& 0& 0 \end{bmatrix}\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}=\begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$

what $x_1,x_2,x_3$ are suitable?

which means $x_1$times column1+$x_2$times column2+$x_3$ times column3=Zero Vector

Look carefully,if you take any linear combination of columns 1 and 2 and don't take column 3(means $x_3=0$), the result will be a zero vector.

So, if we have our X=$\begin{bmatrix} \beta\\ \gamma\\ 0 \end{bmatrix}$ where $\beta, \gamma$ are any scalar quantity,

but $x_3$ must be necessarily zero to get zero vector.

Hence, only option (B) and (D) satisfy.

### 7 Comments

Suppose you have a system like

$\begin{bmatrix} 1 &3 &6 \\ 7 & 5& 9\\ 3 &8 & 2 \end{bmatrix}\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}$

this is equivalent to

$x_1\begin{bmatrix} 1\\ 7\\ 3 \end{bmatrix}+x_2\begin{bmatrix} 3\\ 5\\ 8 \end{bmatrix}+x_3\begin{bmatrix} 6\\ 9\\ 2 \end{bmatrix}$

In the question given, you have to produce a zero vector as result. What suitable combinations of $x_1,x_2,x_3$ can you take?

Himanshu3 No, zero vector can’t be eigen vector of a matrix.

https://math.stackexchange.com/questions/990016/can-the-zero-vector-be-an-eigenvector-for-a-matrix

Consider the characteristic equation $(A-λI)x=0$

$\Rightarrow$ $\begin{bmatrix} 0-\lambda & 0 & \alpha \\ 0 & 0-\lambda & 0\\ 0 & 0 & 0-\lambda \end{bmatrix}$ $\begin{bmatrix}x_1\\ x_2\\ x_3 \end{bmatrix}$ = $\begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$

To find the eigen vector of A corresponding to $\lambda$

Put the eigen value of A in the characteristic equation

i.e. $\begin{bmatrix} 0 & 0 & \alpha \\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}$ $\begin{bmatrix}x_1\\ x_2\\ x_3 \end{bmatrix}$ = $\begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$

Now we have

$\alpha$$x_3$=0

$\Rightarrow$$x_3$ = 0

Therefore, any non zero vector with z-component as zero is an eigen vector.

Hence in the given options $B$ and $D$ are correct.

Eigen values are: $0,0,0$

The eigen vector should satisfy the equation: $\alpha$z $=$ $0$

### 10 Comments

here we have the Basic variable as "z" and free variable as "x" and "y".

So, keeping x = t and y = s, and z = $\alpha$

Then vector corresponding to x is $[1, 0, 0]^T$

And vector corresponding to y is $[0, 1, 0]^T$

Then why it is not $[1, 0, 0]^T$ And $[0, 1, 0]^T$ instead of $[\alpha, 0 , 0]^T$ And $[0, \alpha, 0]^T$?

Guys this is an easy question, so let us not make it complex.

As most of the people have already explained that $|A-\lambda I|*eigenvector =\begin{bmatrix} 0 &0 &\alpha \\ 0 & 0 &0 \\ 0& 0 & 0 \end{bmatrix}$$\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3} \end{bmatrix}$

Now see that we are getting $0*x_{1} + 0*x_{2} + \alpha *x_{3}=0$ as the only useful equation.

Now put whatever the value of $x_{1}$ ,$x_{2}$ and $x_{3}$, we will get our eigenvectors, **EXCEPT **the cases where $\alpha$ has to be zero because in question it clearly mentioned that $\alpha \neq 0$. This condition will only fail when $x_{3}\neq 0.$ And therefore now you can match the options very easily.

clearly matrix is upper triangular matrix and for upper triangular ,lower triangular matrix ,diagonal matrix and scaler matrix ,its diagonal element is eigen values.so here eigen value=0,0,0.

now we have to find eigen vector of this matrix (A-**λ)**X=0 where λ=0 so we get A matrix [ 0 0 α] has rank=1 means we get two free variable (n-r)

x1= β x2=γ x3=0 then eigen vector (x1,x2,x3)=(β,γ,0)= β(1,0,0)+γ(0,1,0) so B and D are right answer .